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Ultraviolet Catastrophe in simple language

  1. Sep 12, 2014 #1
    I am trying to understand what exactly the problem was that drove Plank to develop the quantum scale. I have read wiki about ultraviolet catastrophe and blackbody radiation, but I don't think I really understand what the problem was. Is sounds like the basic problem was that the wavelength of radiation from a substance is not proportional to the amount of energy it contains. What I don't understand, is why it would be expected to be. I would think that the energy level of a substance would be equivalent to the magnitude of a electromagnetic radiation and that the wavelength wouldn't be relevant to the energy levels.

    Thanks
     
  2. jcsd
  3. Sep 12, 2014 #2

    vanhees71

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    Hopefully you haven't read the Wikipedia article on the Ultraviolet Catastrophe. The section "solution" is utterly wrong. You cannot derive the correct Planck Law from the assumption that photons are classical particles as suggested in this section of the article. The key issue is Planck's way to count the microstates for a given macro state in a very specific way. Nowadays we know that this is the correct counting from the quantum field theory of electromagnetic radiation (QED), which is the only correct way to derive the law from the point of view of modern theory. As we know now, Planck's counting is correct, because of the quantum mechanics of bosonic (quantum) field excitations of a field of integer-valued spin!
     
  4. Sep 12, 2014 #3
    Ok, what I am trying to figure out, is what the actual problem was that Plank was trying to solve.
    In other words, I don't really understand the situation in which quanta needed to be created in order to provide a solution. I think it was the ultraviolet catastrophe, but am not 100%. If it is, the UC doesn't really make sense to me.

    thanks
     
  5. Sep 12, 2014 #4
    I'm new to QM and am trying to start at the beginning..
     
  6. Sep 12, 2014 #5

    quo

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    Any derivation of the Planck's law utilises in fact the classical wave modes.
    Thus your stipulated QED theory is inadequate to resolve this problem.
     
  7. Sep 12, 2014 #6

    Jano L.

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    Planck was concerned with explaining the measured spectrum of thermal radiation based on thermodynamics and electromagnetic theory. He was not concerned with solving the problems of the Rayleigh-Jeans approach. The "ultraviolet-catastrophe" is a term coined much later (by Ehrenfest I think).

    https://encrypted.google.com/search?hl=en&q=Max Planck: The Reluctant Revolutionary
     
  8. Sep 12, 2014 #7
    What was anomalous about the measured spectrum of thermal radiation? What needed explained?
     
  9. Sep 12, 2014 #8

    vanhees71

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    The more it is important to learn modern QT right away! It's a lot of effort to understand the "old quantum theory" just to learn later to forget about it. Photons are a pretty complicated subject and thus the worst case to start. Just learn non-relativistic QT first. It's good to begin with the Schrödinger equation and atomic physics of light atoms, where non-relativistic QT is a very good approximation!

    To the contrary! It's the quickest way to derive the Planck spectrum. In QED it reduces to resum a geometric series. In other words, it's a no-brainer!
     
  10. Sep 12, 2014 #9
    Yeah, I have read about modern QT. I'm one of those people that just doesn't understand the need for a quantum scale.. I could continue to memorize formulas taking people's word that a quantum scale is needed, but that just doesn't seem like a good idea. Can anyone here please help me understand why a quantum scale is needed in the first place. Thanks
     
  11. Sep 13, 2014 #10

    quo

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    This is irrelevant.
    Any value can be summed up in infinitely many ways.

    For example:
    1 = 1/2 + 1/4 + 1/8 + 1/16 + ...
    1 = (1/2 + 1/4) + (1/8 + 1/16) + ...
    1 = (1/2 + 1/8) + (1/4 + 1/16) + ...

    ect.
     
  12. Sep 13, 2014 #11

    bhobba

    Staff: Mentor

    You need to study - Ballentine - QM - A Modern Development
    https://www.amazon.com/Quantum-Mechanics-Modern-Development-Edition/dp/9814578584/ref=dp_ob_title_bk

    Chapter 3 will make everything clear.

    Regarding QFT I am studying a book right now that I am really taken with:
    https://www.amazon.com/Quantum-Field-Theory-Gifted-Amateur/dp/019969933X

    It really explains, in very clear language, exactly what's going on.

    Also you do not need advanced QM - an intermediate book like Griffiths would be just fine - even the treatment is general physics texts like University Physics would be adequate.

    And the Kindle price is good for a QFT Book.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  13. Sep 13, 2014 #12

    bhobba

    Staff: Mentor

    Last edited by a moderator: May 6, 2017
  14. Sep 13, 2014 #13

    vanhees71

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    The Planck Law is really easily derived. In the following I use natural units with [itex]\hbar=c=k_{\text{B}}=1[/itex].

    You start from the quantization of the free electromagnetic field (most simply in the radiation gauge, [itex]A^0=0[/itex], [itex]\vec{\nabla} \cdot \vec{A}=0[/itex]). The Fock basis is given by
    [tex]|\{N(\vec{p},\lambda) \} \rangle,[/tex]
    where I use a cube of length [itex]L[/itex] as a quantization volume with periodic boundary conditions. Then [itex]\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3[/itex] and [itex]\lambda \in \{-1,1\}[/itex] (helicities).

    Since photons have spin 1, they are bosons and thus the occupation numbers all run from 0 to [itex]\infty[/itex]. The canonical partition sum for photons of fixed energy [itex]E=|\vec{p}|[/itex] thus is
    [tex]Z(\beta,E)=\sum_{N=0}^{\infty} \exp(-2 \beta E N)=\left (\frac{1}{1-\exp(-\beta E)} \right )^2,[/tex]
    where the square comes from the two helicities (polarizations) of each photon mode.

    The average total energy of the photons of energy [itex]E[/itex] is thus
    [tex]\langle E_{\text{tot}} \rangle_{E}=-\partial_{\beta} \ln Z(\beta,E)=\frac{2 E}{\exp(\beta E)-1}.[/tex]
    To get the energy spectrum, we have to count the states. In a momentum-volume element [itex]\Delta^3 \vec{p}[/itex] we have [itex]\mathrm{\Delta}^3 \vec{p} L^3/(2 \pi)^3[/itex] states. Thus the energy-density spectrum is in the limit [itex]L \rightarrow \infty[/itex]
    [tex]\mathrm{d} u(E)=\frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \frac{2 |\vec{p}|}{\exp(\beta |\vec{p}|)-1}.[/tex]
    Because of [itex]E^2=\vec{p}^2[/itex] we have
    [tex]\mathrm{d}^3 \vec{p} = \vec{p}^2 \mathrm{d} |\vec{p}| \mathrm{d} \Omega= E^2 \mathrm{d} E \mathrm{d} \Omega.[/tex]
    Integrating over the full solid angle gives
    [tex]\mathrm{d} u=\mathrm{d} E \frac{8 \pi}{(2 \pi)^3} \frac{E^3}{\exp(\beta E)-1}.[/tex]
    This is Planck's Law:
    [tex]u(E)=frac{8 \pi}{(2 \pi)^3} \frac{E^3}{\exp(\beta E)-1}.[/tex]
    Usually it's written in terms of the frequency [itex]\nu=E/(2 \pi)[/itex] which gives
    [tex]u(\nu)=2 \pi u(E)=\frac{16 \pi^2 \nu^3}{\exp(2 \pi \beta \nu)-1}[/tex]
    or reinstalling all factors [itex]h=2 \pi \hbar[/itex], [itex]c[/itex] and [itex]k_{\text{B}}[/itex]
    [tex]u(\nu)=\frac{8 \pi h \nu^3}{c^3} \frac{1}{\exp[h \nu/(k_{\text{B}} T)]-1}.[/tex]
    See Wikipedia for a thorough further discussion:

    https://en.wikipedia.org/wiki/Planck's_law
     
  15. Sep 14, 2014 #14

    quo

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    You probably rewrite the old Bose's idea only, which is quite artificial, without any ground in the physics.
    The two modes you called as helicities, are in fact the two polarisation of wave: 0 or 180 - the boundary condition.
     
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